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Engineering Calculus Notes 277

# Engineering Calculus Notes 277 - y giving a point on the...

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3.4. LEVEL CURVES 265 As an illustration of this last point, we again consider the function f ( x,y ) = x 2 + y 2 . The level set L ( f, 1) is the circle of radius 1 about the origin x 2 + y 2 + 1 . We can solve this equation for y in terms of x on any open arc which does not include either of the points ( ± 1 , 0): if the point ( x 0 ,y 0 ) with | x 0 | < 1 has y 0 > 0, the solution near ( x 0 ,y 0 ) is φ ( x ) = radicalbig 1 x 2 whereas if y 0 < 0 it is φ ( x ) = radicalbig 1 x 2 . Since ∂f ∂y = 2 y, at the two points ( ± 1 , 0) ∂f ∂y ( ± 1 , 0) = 0 and the theorem does not guarantee the possibility of solving for y in terms of x ; in fact, for x near ± 1 there are two values of
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Unformatted text preview: y giving a point on the curve, given by the two formulas above. However, since at these points ∂f ∂x ( ± 1 , 0) = ± 2 n = 0 , the theorem does guarantee a solution for x in terms of y ; in fact, near any point other than (0 , ± 1) (the “north pole” and “south pole”) we can write x = ψ ( y ), where ψ ( y ) = r 1 − y 2 for points on the right semicircle and ψ ( y ) = − r 1 − y 2 on the left semicircle....
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